PROBLEM C : 2870 from PROBLEMS for the ANNUAL COMPETITION of "GAZETA MATEMATICĂ"

          In GMB 5/2005 magazine, pages 234 (Romanian version) and 236 (English version).

          The problem says : 

     C:2870.  Find the minimum value of :  $E(x)=\frac{x^2-2x-1}{x^2-2x+3}$,  for  $x\in\mathbb{R}$.

[Author :] Vasile PREDAN, Curtea de Argeș

ANSWER CiP

$-1=E(1)\leqslant E(x)<1$

                     Solution CiP  We will imitate the solution from the Post here.

           The number  $\lambda$  is a value of  $E(x)\;\;\Leftrightarrow$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;E(x)=\lambda$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;x^2-2x-1=\lambda \cdot (x^2-2x+3)$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;(1-\lambda)x^2+2(\lambda-1)x-(3\lambda+1)=0 \tag{1}$

$\Leftrightarrow$  the quadratic equation (1) has solution(s).

[For  $\lambda=1$  equation (1) isn't quadratic both then in hasn't solution.].

     Using the half discriminant ($\Delta'=(b')^2-ac$) formula, the roots of equation (1) are

$x_{1,2}=\frac{1-\lambda\pm\sqrt{2-2\lambda^2}}{1-\lambda} \tag{2}$

The requirement that the roots be real numbers, $|\Delta'\geqslant 0$  implies  $2-2\lambda^2\geqslant 0\;\Leftrightarrow$

$\Leftrightarrow\;\lambda^2\leqslant 1\;\Leftrightarrow\;-1\leqslant \lambda<1.$ From (2) we obtain the expressions of the roots as follows  $x_{1,2}=1\pm \sqrt{\frac{2(1+\lambda)}{1-\lambda}}.$

$\blacksquare$

       Writing differently,  $E(x)=1-\frac{4}{(x-1)^2+2}\;\;\Rightarrow\;\; -1=1-\frac{4}{0+2}\leqslant 1-\frac{4}{(x-1)^2+2}<1$,  with the same answer.